B.Crowell - Conservation Laws, Vol

Algebraic Proof of the Result in the Example

The equation A+B = C+D says that the change in one ball’s velocity is equal and opposite to the change in the other’s. We invent a symbol x=C-A for the change in ball 1’s velocity. The second equation can then be rewritten as

A2+B2 = (A+x)2+(B-x)2 .

Squaring out the quantities in parentheses gives

A2+B2

= A2+2Ax+x2+B2-2Bx+x2 ,

which simplifies to

0 = Ax-Bx+x2 .

The equation has the trivial solution x=0, i.e. neither ball’s velocity is changed, but this is physically impossible because the balls cannot travel through each other like ghosts. Assuming x¹0, we can divide by x and solve for x=B-A. This means that ball 1 has gained an amount of velocity exactly sufficient to make it equal to ball 2’s initial velocity, and vice-versa. The balls must have swapped velocities.

while it was being squashed by the impact.

Collisions of the superball type, in which almost no kinetic energy is converted to other forms of energy, can thus be analyzed more thoroughly, because they have KEf=KEi, as opposed to the less useful inequality KEf<KEi for a case like a tennis ball bouncing on grass.

Example: pool balls colliding head-on

Question: Two pool balls collide head-on, so that the collision is restricted to one dimension. Pool balls are constructed so as to lose as little kinetic energy as possible in a collision, so under the assumption that no kinetic energy is converted to any other form of energy, what can we predict about the results of such a collision?

Solution: Pool balls have identical masses, so we use the same symbol m for both. Conservation of energy and no loss of kinetic energy give us the two equations

mv1i + mv2i = mv1f + mv2f

12mv1i2 + 12mv2i2 = 12mv1f2 + 12mv2f2

The masses and the factors of 1/2 can be divided out, and we eliminate the cumbersome subscripts by replacing the symbols v1i,... with the symbols A, B, C, and D.

A+B = C+D A2+B2 = C2+D2 .

A little experimentation with numbers shows that given values of A and B, it is impossible to find C and D that satisfy these equations unless C and D equal A and B, or C and D are the same as A and B but swapped around. An algebraic proof is given in the box on the left. In the special case where ball 2 is initially at rest, this tells us that ball 1 is stopped dead by the collision, and ball 2 heads off at the velocity originally possessed by ball 1. This behavior will be familiar to players of pool.

Often, as in the example above, the details of the algebra are the least interesting part of the problem, and considerable physical insight can be gained simply by counting the number of unknowns and comparing to the number of equations. Suppose a beginner at pool notices a case where her cue ball hits an initially stationary ball and stops dead. “Wow, what a good trick,” she thinks. “I bet I could never do that again in a million years.” But she tries again, and finds that she can’t help doing it even if she doesn’t want to. Luckily she has just learned about collisions in her physics course. Once she has written down the equations for conservation of energy and no loss of kinetic energy, she really doesn’t have to complete the algebra. She knows that she has two equations in two unknowns, so there must be a welldefined solution. Once she has seen the result of one such collision, she knows that the same thing must happen every time. The same thing would happen with colliding marbles or croquet balls. It doesn’t matter if the masses or velocities are different, because that just multiplies both equations by some constant factor.

Chadwick’s subatomic pool table. A disk of the naturally occurring metal polonium provides a source of radiation capable of kicking neutrons out of the beryllium nuclei. The type of radiation emitted by the polonium is easily absorbed by a few mm of air, so the air has to be pumped out of the left-hand chamber. The neutrons, Chadwick’s mystery particles, penetrate matter far more readily, and fly out through the wall and into the chamber on the right, which is filled with nitrogen or hydrogen gas. When a neutron collides with a nitrogen or hydrogen nucleus, it kicks it out of its atom at high speed, and this recoiling nucleus then rips apart thousands of other atoms of the gas. The result is an electrical pulse that can be detected in the wire on the right. Physicists had already calibrated this type of apparatus so that they could translate the strength of the electrical pulse into the velocity of the recoiling nucleus. The whole apparatus shown in the figure would fit in the palm of your hand, in dramatic contrast to today’s giant particle accelerators.

to vacuum pump

polonium beryllium

In this multiple-flash photograph, we see the wrench from above as it flies through the air, rotating as it goes. Its center of mass, marked with the black cross, travels along a straight line, unlike the other points on the wrench, which execute loops.

Two hockey pucks collide. Their mutual center of mass traces the straight path shown by the dashed line.

We have already discussed the idea of the center of mass in the first book of this series, but using the concept of momentum we can now find a mathematical method for defining the center of mass, explain why the motion of an object’s center of mass usually exhibits simpler motion than any other point, and gain a very simple and powerful way of understanding collisions.

The first step is to realize that the center of mass concept can be applied to systems containing more than one object. Even something like a wrench, which we think of as one object, is really made of many atoms. The center of mass is particularly easy to visualize in the case shown on the left, where two identical hockey pucks collide. It is clear on grounds of symmetry that their center of mass must be at the midpoint between them. After all, we previously defined the center of mass as the balance point, and if the two hockey pucks were joined with a very lightweight rod whose own mass was negligible, they would obviously balance at the midpoint. It doesn’t matter that the hockey pucks are two separate objects. It is still true that the motion of their center of mass is exceptionally simple, just like that of the wrench’s center of mass.

The x coordinate of the hockey pucks’ center of mass is thus given by xcm=(x1+x2)/2, i.e. the arithmetic average of their x coordinates. Why is its motion so simple? It has to do with conservation of momentum. Since the hockey pucks are not being acted on by any net external force, they constitute a closed system, and their total momentum is conserved. Their total momentum is

= mtotalvcm,x

In other words, the total momentum of the system is the same as if all its

Moving our head so that we are always looking down from right above the center of mass, we observe the collision of the two hockey pucks in the center of mass frame.

mass was concentrated at the center of mass point. Since the total momentum is conserved, the x component of the center of mass’s velocity vector cannot change. The same is also true for the other components, so the center of mass must move along a straight line at constant speed.

The above relationship between the total momentum and the motion of the center of mass applies to any system, even if it is not closed.

total momentum related to center of mass motion

The total momentum of any system is related to its total mass and the velocity of its center of mass by the equation

ptotal = mtotal vcm .

What about a system containing objects with unequal masses, or containing more than two objects? The reasoning above can be generalized to a weighted average

with similar equations for the y and z coordinates.

Momentum in different frames of reference

Absolute motion is supposed to be undetectable, i.e. the laws of physics are supposed to be equally valid in all inertial frames of reference. If we first calculate some momenta in one frame of reference and find that momentum is conserved, and then rework the whole problem in some other frame of reference that is moving with respect to the first, the numerical values of the momenta will all be different. Even so, momentum will still be conserved. All that matters is that we work a single problem in one consistent frame of reference.

One way of proving this is to apply the equation ptotal=mtotalvcm. If the velocity of frame B relative to frame A is vBA, then the only effect of chang-

ing frames of reference is to change vcm from its original value to vcm+vBA. This adds a constant onto the momentum vector, which has no effect on conservation of momentum.

The center of mass frame of reference

A particularly useful frame of reference in many cases is the frame that moves along with the center of mass, called the center of mass (c.m.) frame. In this frame, the total momentum is zero. The following examples show how the center of mass frame can be a powerful tool for simplifying our understanding of collisions.

Example: a collision of pool balls viewed in the c.m. frame

If you move your head so that your eye is always above the point halfway in between the two pool balls, you are viewing things in the center of mass frame. In this frame, the balls come toward the center of mass at equal speeds. By symmetry, they must therefore recoil at equal speeds along the lines on which they entered. Since the balls have essentially swapped paths in the center of mass frame, the same must also be true in any other frame. This is the same result that required laborious algebra to prove previously without the concept of the center of mass frame.

v1f

v2

v1i

(a) The slingshot effect viewed in the sun’s frame of reference. Jupiter is moving to the left, and the collision is head-on.

v1f

v1i

(b) The slingshot viewed in the frame of the center of mass of the Jupiterspacecraft system.

Example: the slingshot effect

It is a counterintuitive fact that a spacecraft can pick up speed by swinging around a planet, if arrives in the opposite direction compared to the planet’s motion. Although there is no physical contact, we treat the encounter as a one-dimensional collision, and analyze it in the center of mass frame. Since Jupiter is so much more massive than the spacecraft, the center of mass is essentially fixed at Jupiter’s center, and Jupiter has zero velocity in the center of mass frame, as shown in figure (b). The c.m. frame is moving to the left compared to the sun-fixed frame used in (a), so the spacecraft’s initial velocity is greater in this frame. Things are simpler in the center of mass frame, because it is more symmetric. In the sun-fixed frame, the incoming leg of the encounter is rapid, because the two bodies are rushing toward each other, while their separation on the outbound leg is more gradual, because Jupiter is trying to catch up. In the c.m. frame, Jupiter is sitting still, and there is perfect symmetry between the incoming and outgoing legs, so by symmetry we have v1f=−v1i. Going back to the sun-fixed frame, the spacecraft’s final velocity is increased by the frames’ motion relative to each other. In the sun-fixed frame, the spacecraft’s velocity has increased greatly. The result can also be understood in terms of work and energy. In Jupiter’s frame, Jupiter is not doing any work on the spacecraft as it rounds the back of the planet, because the motion is perpendicular to the force. But in the sun’s frame, the spacecraft’s velocity vector at the same moment has a large component to the left, so Jupiter is doing work on it.

energy

power = rate of

transferring energy

system

momentum

force = rate of transferring momentum

Thus the rate of transfer of momentum, i.e. the number of kg.m/s absorbed per second, is simply the external force,

[ relationship between the force on an object and the rate of change of its momentum; valid only if the force is constant ]

This equation is really just a restatement of Newton’s second law, and in fact Newton originally stated it this way. As shown in the diagram, the relationship between force and momentum is directly analogous to that between power and energy.

The situation is not materially altered for a system composed of many objects. There may be forces between the objects, but the internal forces cannot change the system’s momentum — if they did, then removing the external forces would result in a closed system that was able to change its own momentum, violating conservation of momentum. The equation above becomes

Example: walking into a lamppost

Question: Starting from rest, you begin walking, bringing your momentum up to 100 kg.m/s. You walk straight into a lamppost. Why is the momentum change of -100 kg.m/s so much more painful than the change of +100 kg.m/s when you started walking?

Solution: The situation is one-dimensional, so we can dispense with the vector notation. It probably takes you about 1 s to speed up initially, so the ground’s force on you is F= p/ t≈100 N. Your impact with the lamppost, however, is over in the blink of an eye, say 1/10 s or less. Dividing by this much smaller t gives a much larger force, perhaps thousands of newtons. (The negative sign simply indicates that the force is in the opposite direction.)

This is also the principle of airbags in cars. The time required for the airbag to decelerate your head is fairly long, the time required for your face to travel 20 or 30 cm. Without an airbag, your face would have been hitting the dashboard, and the time interval would have been the much shorter time taken by your skull to move a couple of centimeters while your face compressed. Note that either way, the same amount of mechanical work has to be done on your head: enough to eliminate all its kinetic energy.